comparing procedures

[Gregory Marton]

Hi folks,

I'm trying to write a meaningful comparison operator for procedures. 
Clearly this wants more than procedure-source, because variables in the 
source may be bound to different values in the procedure-environment.

I expected something like this to work:
(define foo 3)
(define bar (lambda (x) (+ x foo)))
(define baz (lambda (x) (* x foo)))
(define bar-env (procedure-environment bar))
(define baz-env (procedure-environment baz))
(environment-fold
  bar-env
  (lambda (sym val so-far)
    (and so-far
         (environment-bound? baz-env sym)
         (equal? val (environment-ref baz-env sym))))
  #t)

But it turns out that procedure-environment returns something which is not 
something these procedures take as an argument, it's an eval-closure.
(environment? bar-env) ===> #f
(environment? (car bar-env)) ===> #f

I would have thought that this was the wrong procedure but for postings 
like this:
   http://www.cs.brown.edu/pipermail/plt-scheme/2005-August/009540.html

Could someone point me towards the right way to compare the environment of 
a procedure field by field?  (Yes, I also have to make sure there are no 
extra fields in baz-env too, but that part I'll figure out.)

I did dig a little farther, because I can still do this:
(local-eval 'foo bar-env) ===> 3
and local-eval says:
  -- Scheme Procedure: local-eval exp [env]
      Evaluate EXP in its environment.  If ENV is supplied, it is the
      environment in which to evaluate EXP.  Otherwise, EXP must be a
      memoized code object (in which case, its environment is implicit).

But perhaps not all environments are created equal?

This is in 1.8.1 and 1.8.3.

Thanks, confusedly,
Grem

--
------ __@   Gregory A. Marton                http://csail.mit.edu/~gremio/
--- _`\<,_                                                                .
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